Problem: Simplify the following expression: $y = \dfrac{2x^2- 15x+28}{2x - 7}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(28)} &=& 56 \\ {a} + {b} &=& &=& {-15} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $56$ and add them together. The factors that add up to ${-15}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-7}$ and ${b}$ is ${-8}$ $ \begin{eqnarray} {ab} &=& ({-7})({-8}) &=& 56 \\ {a} + {b} &=& {-7} + {-8} &=& -15 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 {-7}x) + ({-8}x +{28}) $ Factor out the common factors: $ x(2x - 7) - 4(2x - 7)$ Now factor out $(2x - 7)$ $ (2x - 7)(x - 4)$ The original expression can therefore be written: $ \dfrac{(2x - 7)(x - 4)}{2x - 7}$ We are dividing by $2x - 7$ , so $2x - 7 \neq 0$ Therefore, $x \neq \frac{7}{2}$ This leaves us with $x - 4; x \neq \frac{7}{2}$.